Question

A block resting on the horizontal surface executes SHM in horizontal plane with amplitude $A$. The frequency of oscillation for which the block just starts to slip is
($\mu =$ coefficient of friction, $g=$ gravitational acceleration)

• A. $\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$
• B. $\frac{1}{4\pi }\sqrt{\frac{\mu g}{A}}$
• C. $2\pi \sqrt{\frac{A}{\mu g}}$
• D. $4\pi \sqrt{\frac{A}{\mu g}}$

Answer 1

The correct option is A $\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$

When restoring force will become equal to the frictional force, block will start to slip.
$\therefore$ Restoring force = Friction force
$\Rightarrow kA=\mu mg$ ....(i)
Now, frequency $f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
From equation (i)
$f=\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$